Question

determine as raízes abaixo calculando sequentemente Bhaskara.
a) (x+2)^2=9

Urgente!

Answer 1

$(x+2)^2=9 \\ x^2+4x+4=9 \\ x^2+4x+4-9=0 \\ x^2+4x-5=0 \\ \\ a=1 \\ b=4 \\ c=-5 \\ \\ \Delta=b^2-4ac \\ \Delta=4^2-4(1)(-5) \\ \Delta=16+25 \\ \Delta=36 \\ \\ x= \frac{-b\pm \sqrt{\Delta} }{2a} = \frac{-4\pm \sqrt{36} }{2} = \frac{-4\pm6}{2} \\ \\ x'= \frac{-4+6}{2} = \frac{2}{2} =1 \\ \\ x"= \frac{-4-6}{2} =- \frac{10}{2} =-5$